If ${\tan }^{-1}\left(\frac{2\mathrm{x}}{{\mathrm{x}}^2-1}\right)+{\cos }^{-1}\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)=\frac{2\mathrm{\pi }}{3}$ then x=

$$\begin{gathered} \mathrm{Put} \mathrm{x}=\mathrm{Tan\theta } \Rightarrow {\tan }^{-1}\left(\frac{2\mathrm{x}}{{\mathrm{x}}^2-1}\right)={\tan }^{-1}\left(-\left(\frac{2 \tan \mathrm{\theta }}{1-{\tan }^2 \mathrm{\theta }}\right)\right)=-{\tan }^{-1}\left(\tan 2\mathrm{\theta }\right) \\ and {\cos }^{-1}\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)={\cos }^{-1}\left(-\left(\frac{1-{\tan }^2\mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}\right)\right)={\cos }^{-1}\left(-\cos 2\theta \right)=\mathrm{\pi }-{\cos }^{-1}\left(\cos 2\mathrm{\theta }\right) \\ {\tan }^{-1}\left(\frac{2\mathrm{x}}{{\mathrm{x}}^2-1}\right)+ {\cos }^{-1}\left(\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2+1}\right)=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow -{\tan }^{-1}\left(\tan 2\mathrm{\theta }\right)+\mathrm{\pi }-{\cos }^{-1}\left(\cos 2\mathrm{\theta }\right)=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow -2\mathrm{\theta }+\mathrm{\pi }-2\mathrm{\theta }=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow \mathrm{\pi }-4\mathrm{\theta }=\frac{2\mathrm{\pi }}{3}\therefore 4\mathrm{\theta }=\frac{\mathrm{\pi }}{3}\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{12} \\ \therefore \mathrm{x}=\tan \frac{\mathrm{\pi }}{12}\Rightarrow \mathrm{x}=2-\sqrt{3} \end{gathered}$$